AQA A-Level Chemistry Paper 1, 2019: Question 2

10 marks · Medium difficulty · State/Explain/Numerical

Describe electrospray ionisation, deduce relative molecular mass from a mass spectrum peak, and calculate the relative molecular mass of a molecule using time of flight data.

Practise this question

Question

Question 02 begins with a TOF mass spectrum showing a single peak at m/z = 556 for pentapeptide P. Question 02.1 asks to describe electrospray ionisation and provide an equation for ionisation of P (4 marks). Question 02.2 asks to choose the relative molecular mass of P from 555, 556, or 557 (1 mark). Question 02.3 presents molecule Q ionised by electron impact with KE = 2.09 x 10^-15 J, time of flight 1.23 x 10^-5 s, flight tube length 1.50 m, and asks to calculate the relative molecular mass using KE = 1/2 mv^2 and the Avogadro constant (5 marks).
Question text

02 Time of flight (TOF) mass spectrometry can be used to analyse large molecules such

as the pentapeptide, leucine encephalin (P).

P is ionised by electrospray ionisation and its mass spectrum is shown in Figure 2.

Figure 2

02.1 Describe the process of electrospray ionisation.

Give an equation to represent the ionisation of P in this process.

[4 marks]

Description

Equation

02.2 What is the relative molecular mass of P?

Tick ( ) one box.

[1 mark]

555 556 557

02.3 A molecule Q is ionised by electron impact in a TOF mass spectrometer.

The Q+ ion has a kinetic energy of 2.09 x 10–15 J

This ion takes 1.23 x 10–5 s to reach the detector.

The length of the flight tube is 1.50 m

Calculate the relative molecular mass of Q.

12 –1

KE = mv where m = mass (kg) and v = speed (m s )

The Avogadro constant, L = 6.022 x 1023 mol–1

[5 marks]

Relative molecular mass

Mark scheme

Show the mark scheme Mark scheme for Question 02. 02.1 awards 4 marks: P dissolved in a solvent, injected through a needle at high voltage/potential, gains a proton/H+, and equation P + H+ -> PH+. 02.2 awards 1 mark for 555. 02.3 awards 5 marks: M1 for calculating velocity v = d/t = 1.22 x 10^5 m/s, M2 for rearranging to find mass m = 2KE/v^2, M3 for mass of one ion m = 2.81 x 10^-25 kg, M4 for multiplying by Avogadro constant L = 0.169 kg, and M5 for multiplying by 1000 to get relative molecular mass = 169.

Question Answers Additional Comments/Guidelines Mark

M1: P dissolved or put in/added to a solvent M1: Allow named solvent eg water or methanol 1

M2: (injected through) a needle or nozzle or capillary and at high M2: Allow needle is positively charged 1

voltage/4000 volts or high potential

M3: Gains a proton / H+

02.1 M3: Not atoms gain a proton 1

M3: Could be scored from equation

M4: P + H+ PH+

Correct equation gains M3 and M4 1

Ignore state symbols

02.2 555 1

M1 V = d/t or = 1.22 x 105 ms-1 Recall this equation

M2 m = 2KE or 2 x 2.09 x 10–15

Rearrangement to give m 1

v2 (1.22 x 105)2

or

M2 m = 2KE x t2 or 2 x 2.09 x 10–15 x (1.23 x 10 –5)2

d2 1.502

02.3 -25

M3 m = 2.8(1) x 10 (kg) M3: Calculation of m. 1

M4 = 2.81 x 10-25 x L = 0.169

M4: Allow M3 x L 1

M5 0.169 x 1000 = 169.(2) M5: Allow M4 x 1000

169 only scores 5 marks 1

Allow answers to 2 significant figures or more

ignore units

How to answer it

TOF Mass Spectrometry: Ionisation & Calculations

📌 What this question tests

This question examines core knowledge of Time of Flight (TOF) mass spectrometry: describing the 4-step mechanism of electrospray ionisation, deducing molecular mass from an electrospray mass spectrum ([M + H]⁺ ion peak), and performing multi-step kinetic energy calculations to determine the relative molecular mass (Mr) of an ionised particle using flight times, distance, and the Avogadro constant.

Question 02.1 (4 Marks)

Electrospray Ionisation Mechanism & Equation

Describing soft ionisation and writing the ionisation equation for substance P

✅ Model Answer & Mark Breakdown

  • M1: Substance P is dissolved in a volatile solvent (e.g. water or methanol).
  • M2: Injected through a fine hypodermic needle/nozzle attached to a high voltage supply (or high potential).
  • M3: Each molecule gains a proton (H⁺) from the solvent.
  • M4 (Equation): P + H⁺ → PH⁺ (or P(g) + H⁺(g) → PH⁺(g))
ℹ️ A correct equation alone automatically awards both M3 and M4. State symbols are not required.

💡 Key Knowledge

  • Electrospray vs Electron Impact: Electrospray is a "soft" ionisation technique used for large biological molecules (like pentapeptides) because it prevents fragmentation.
  • The particle detected is an adduct ion: [M + H]⁺ , having a mass of Mr + 1.
  • The solvent evaporates, leaving positive ions moving towards a negative plate.

🧠 Exam Technique

Learn the 3 distinct prose points for electrospray by heart:

  1. Dissolve in volatile solvent.
  2. Inject through needle at high voltage.
  3. Gain a proton (H⁺).

Always write the ion as PH⁺ or [P+H]⁺ , never just P⁺ .

❌ Common Errors

  • Writing that "an electron is knocked off" – this confuses electrospray with electron impact!
  • Stating that "atoms gain a proton" – leucine encephalin is a molecule, not an atom.
  • Forgetting to state the needle requires a high voltage (simply saying "sprayed from a needle" scores 0 for M2).
Question 02.2 (1 Mark)

Deducing Relative Molecular Mass (Mr)

Interpreting the peak produced by electrospray ionisation

✅ Correct Answer

Select: 555

Tick box 1: [✔] 555    [   ] 556    [   ] 557

💡 Key Knowledge & Examiner Insight

  • The mass spectrum shows a single peak at m/z = 556 .
  • Because electrospray adds a proton (H⁺), the ion detected is PH⁺ :
    m/z = Mr + 1 = 556
  • Therefore: Mr = 556 − 1 = 555 .
⚠️ Common Trap: Ticking 556. Many students forget that electrospray ionisation adds an H⁺ ion (mass 1), unlike electron impact which retains the original molecule's mass number.
Question 02.3 (5 Marks)

TOF Flight Tube Calculations

Calculating the relative molecular mass (Mr) of molecule Q

📐 Step-by-Step Calculation

Given Data:

  • Kinetic energy, KE = 2.09 × 10⁻¹⁵ J
  • Flight time, t = 1.23 × 10⁻⁵ s
  • Flight path length, d = 1.50 m
  • Avogadro constant, L = 6.022 × 10²³ mol⁻¹

1 Calculate velocity (v) of ion Q⁺:

v = d / t = 1.50 / (1.23 × 10⁻⁵) = 1.2195 × 10⁵ m s⁻¹
awarded M1

2 Rearrange KE equation for mass (m) of one ion:

KE = ½mv²   ➜   m = 2(KE) / v²
(or combine into m = 2(KE) × t² / d² )
awarded M2

3 Calculate mass of a single ion in kg:

m = [2 × (2.09 × 10⁻¹⁵)] / (1.2195 × 10⁵)² = 2.8105 × 10⁻²⁵ kg
awarded M3 (Allow 2.8 × 10⁻²⁵ to 2.81 × 10⁻²⁵ kg)

4 Calculate mass of 1 mole of ions in kg:

mmol = m × L = (2.8105 × 10⁻²⁵) × (6.022 × 10²³) = 0.16925 kg mol⁻¹
awarded M4

5 Convert mass of 1 mole to grams (Mr):

Relative molecular mass must be expressed in g mol⁻¹:
Mr = 0.16925 × 1000 = 169.2 (or 169)
awarded M5 (Accept 169 to 169.3; 2 or more sig figs accepted, ignore units)

🧠 Exam Technique: Single Equation Shortcut

You can substitute v = d / t directly into KE = ½mv² :

m = 2 · KE · t² / d²

This avoids rounding errors midway through the calculation! Always keep values stored in your calculator memory.

❌ Common Traps & Lost Marks

  • Forgetting × 1000 at the end: The mass from KE = ½mv² is always in kg. Mr corresponds to g mol⁻¹. Forgetting to multiply by 1000 loses M5.
  • Premature rounding: Rounding velocity to 1.2 × 10⁵ too early can cause downstream inaccuracies that lead to a final answer outside the acceptable range.
  • Squaring errors: Forgetting to square v or squaring the wrong term when rearranging.

Topics

Physical Chemistry · 3.1.1 Atomic Structure · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.